What is the limit as x approaches 0+of x[ln(x)]^2 ?

The limit as x approaches 0+of x[ln(x)]^2 is 0

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limit as x approaches 0+of x[ln(x)]^2

x \to 0 + lim (x [ln (x)]^{2})

0

Solution steps

x \to 0 + lim (x [ln (x)]^{2})

Simplify

= x \to 0 + lim (x ln^{2} (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ^{2} ( x )}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}})

= x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}})

Simplify

\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}} : - 2 x ln (x)

= x \to 0 + lim (- 2 x ln (x))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - 2 \cdot x \to 0 + lim (x ln (x))

Rewrite for L'Hopital

= - 2 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

= - 2 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

Simplify

\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}} : - x

= - 2 \cdot x \to 0 + lim (- x)

Plug in the value

x = 0

= - 2 (- 0)

Simplify

= 0

\frac{d}{d x} (2^{5 x} 3^{4 x^{2}})

\frac{d y}{d x} = \frac{x - 4}{y + 5}

f (x) = \frac{2}{x}, (6, \frac{1}{3})

\int 15 ln (x + 5) d x

What is the limit as x approaches 0+of x[ln(x)]^2 ?
The limit as x approaches 0+of x[ln(x)]^2 is 0